UPSC 2025 Maths Optional Paper 2 Q7b — Step-by-Step Solution

15 marks · Section B

Lagrange's interpolation · Numerical Analysis · asked 5× in 13 yrs · Read the full method →

Question

Find the unique polynomial of degree 2 or less which fits the following data:

x : 0, 1, 3 \qquad f(x) : 1, 3, 55

Also obtain the bound on the truncation error.

Technique

Use Newton’s divided-difference interpolation on the three unequally spaced points to get the unique polynomial of degree $\le 2$ . The truncation error uses the standard remainder $f(x) - P_2(x) = \dfrac{f'''(\xi)}{3!}\,\omega(x)$ with $\omega(x) = (x-x_0)(x-x_1)(x-x_2)$ .

Solution

Data points: $(x_0,f_0) = (0,1)$ , $(x_1,f_1) = (1,3)$ , $(x_2,f_2) = (3,55)$ .

Newton’s divided differences

First order:

f[x_0,x_1] = \frac{3 - 1}{1 - 0} = 2,\qquad f[x_1,x_2] = \frac{55 - 3}{3 - 1} = \frac{52}{2} = 26.

Second order:

f[x_0,x_1,x_2] = \frac{26 - 2}{3 - 0} = \frac{24}{3} = 8.

Newton form

P_2(x) = f_0 + f[x_0,x_1](x - x_0) + f[x_0,x_1,x_2](x - x_0)(x - x_1).

P_2(x) = 1 + 2(x - 0) + 8(x - 0)(x - 1) = 1 + 2x + 8x^2 - 8x.

\boxed{P_2(x) = 8x^2 - 6x + 1.}

Check: $P_2(0)=1$ , $P_2(1)=8-6+1=3$ , $P_2(3)=72-18+1=55$ . ✓

Truncation-error bound

For interpolation at three nodes, the error at a point $x$ is

E(x) = f(x) - P_2(x) = \frac{f'''(\xi)}{3!}\,\omega(x),\qquad \omega(x) = x(x-1)(x-3),

for some $\xi$ in the smallest interval containing $\{0,1,3,x\}$ . Hence

|E(x)| \le \frac{M_3}{6}\,|\omega(x)|,\qquad M_3 = \max_{[0,3]}|f'''(\xi)|.

To bound $|\omega(x)|$ on $[0,3]$ : $\omega'(x) = 3x^2 - 8x + 3 = 0$ gives $x = \dfrac{8 \pm \sqrt{64-36}}{6} = \dfrac{4 \pm \sqrt7}{3}$ , i.e. $x \approx 0.4514,\,2.2153$ . Evaluating:

|\omega(0.4514)| \approx 0.6311,\qquad |\omega(2.2153)| \approx 2.1126,

and $\omega = 0$ at the nodes. So

\max_{[0,3]}|\omega(x)| = |\omega(2.2153)| \approx 2.1126\quad\Big(\text{exactly at } x = \tfrac{4+\sqrt7}{3}\Big).

Therefore the uniform bound on the truncation error over $[0,3]$ is

\boxed{\,|f(x) - P_2(x)| \le \frac{M_3}{6}\,\max_{[0,3]}|\omega(x)| \approx \frac{M_3}{6}(2.1126) \approx 0.3521\,M_3,\quad M_3 = \max_{[0,3]}|f'''|.}

(If the underlying $f$ is itself a polynomial of degree $\le 2$ , then $f''' \equiv 0$ and the interpolation is exact.)

Answer

$P_2(x) = 8x^2 - 6x + 1$ . Truncation error $E(x) = \dfrac{f'''(\xi)}{6}\,x(x-1)(x-3)$ , bounded on $[0,3]$ by $\dfrac{M_3}{6}\max|\omega| \approx 0.3521\,M_3$ , the maximum of $|\omega|$ occurring at $x = \tfrac{4+\sqrt7}{3}$ .